Kitabı oku: «The Canterbury Puzzles, and Other Curious Problems», sayfa 12
46.—The Riddle of the Crusaders
The correct answer is that there would have been 602,176 Crusaders, who could form themselves into a square 776 by 776; and after the stranger joined their ranks, they could form 113 squares of 5,329 men—that is, 73 by 73. Or 113 × 732 – 1 = 7762. This is a particular case of the so-called "Pellian Equation," respecting which see A. in M., p. 164.
47.—The Riddle of St. Edmondsbury
The reader is aware that there are prime numbers and composite whole numbers. Now, 1,111,111 cannot be a prime number, because if it were the only possible answers would be those proposed by Brother Benjamin and rejected by Father Peter. Also it cannot have more than two factors, or the answer would be indeterminate. As a matter of fact, 1,111,111 equals 239 x 4649 (both primes), and since each cat killed more mice than there were cats, the answer must be 239 cats. See also the Introduction, p. 18.
Treated generally, this problem consists in finding the factors, if any, of numbers of the form (10n – 1)/9.
Lucas, in his L'Arithmétique Amusante, gives a number of curious tables which he obtained from an arithmetical treatise, called the Talkhys, by Ibn Albanna, an Arabian mathematician and astronomer of the first half of the thirteenth century. In the Paris National Library are several manuscripts dealing with the Talkhys, and a commentary by Alkalaçadi, who died in 1486. Among the tables given by Lucas is one giving all the factors of numbers of the above form up to n = 18. It seems almost inconceivable that Arabians of that date could find the factors where n = 17, as given in my Introduction. But I read Lucas as stating that they are given in Talkhys, though an eminent mathematician reads him differently, and suggests to me that they were discovered by Lucas himself. This can, of course, be settled by an examination of Talkhys, but this has not been possible during the war.
The difficulty lies wholly with those cases where n is a prime number. If n = 2, we get the prime 11. The factors when n = 3, 5, 11, and 13 are respectively (3 . 37), (41 . 271), (21,649 . 513,239), and (53 . 79 . 265371653). I have given in these pages the factors where n = 7 and 17. The factors when n= 19, 23, and 37 are unknown, if there are any.2 When n = 29, the factors are (3,191 . 16,763 . 43,037. 62,003 . 77,843,839,397); when n = 31, one factor is 2,791; and when n = 41, two factors are (83 . 1,231).
As for the even values of n, the following curious series of factors will doubtless interest the reader. The numbers in brackets are primes.
n = 2 = (11)
n = 6 = (11) × 111 × 91
n = 10 = (11) × 11,111 × (9,091)
n = 14 = (11) × 1,111,111 × (909,091)
n = 18 = (11) × 111,111,111 × 90,909,091
Or we may put the factors this way:—
n = 2 = (11)
n = 6 = 111 × 1,001
n = 10 = 11,111 × 100,001
n = 14 = 1,111,111 × 10,000,001
n = 18 = 111,111,111 × 1,000,000,001
In the above two tables n is of the form 4m + 2. When n is of the form 4m the factors may be written down as follows:—
n= 4 = (11) × (101)
n = 8 = (11) × (101) × 10,001
n = 12 = (11) × (101) × 100,010,001
n = 16 = (11) × (101) × 1,000,100,010,001.
When n = 2, we have the prime number 11; when n = 3, the factors are 3 . 37; when n = 6, they are 11 . 3 . 37 . 7. 13; when n = 9, they are 32 . 37 . 333,667. Therefore we know that factors of n = 18 are 11. 32 . 37 . 7 . 13 . 333,667, while the remaining factor is composite and can be split into 19 . 52579. This will show how the working may be simplified when n is not prime.
48.—The Riddle of the Frogs' Ring
The fewest possible moves in which this puzzle can be solved are 118. I will give the complete solution. The black figures on white discs move in the directions of the hands of a clock, and the white figures on black discs the other way. The following are the numbers in the order in which they move. Whether you have to make a simple move or a leaping move will be clear from the position, as you never can have an alternative. The moves enclosed in brackets are to be played five times over: 6, 7, 8, 6, 5, 4, 7, 8, 9, 10, 6, 5, 4, 3, 2, 7, 8, 9, 10, 11 (6, 5, 4, 3, 2, 1), 6, 5, 4, 3, 2, 12, (7, 8, 9, 10, 11, 12), 7, 8, 9, 10, 11, 1, 6, 5, 4, 3, 2, 12, 7, 8, 9, 10, 11, 6, 5, 4, 3, 2, 8, 9, 10, 11, 4, 3, 2, 10, 11, 2. We thus have made 118 moves within the conditions, the black frogs have changed places with the white ones, and 1 and 12 are side by side in the positions stipulated.
The general solution in the case of this puzzle is 3n2 + 2n – 2 moves, where the number of frogs of each colour is n. The law governing the sequence of moves is easily discovered by an examination of the simpler cases, where n = 2, 3, and 4.
If, instead of 11 and 12 changing places, the 6 and 7 must interchange, the expression is n2 + 4n + 2 moves. If we give n the value 6, as in the example of the Frogs' Ring, the number of moves would be 62.
For a general solution of the case where frogs of one colour reverse their order, leaving the blank space in the same position, and each frog is allowed to be moved in either direction (leaping, of course, over his own colour), see "The Grasshopper Puzzle" in A. in M., p. 193.
THE STRANGE ESCAPE OF THE KING'S JESTER
Although the king's jester promised that he would "thereafter make the manner thereof plain to all," there is no record of his having ever done so. I will therefore submit to the reader my own views as to the probable solutions to the mysteries involved.
49.—The Mysterious Rope
When the jester "divided his rope in half," it does not follow that he cut it into two parts, each half the original length of the rope. No doubt he simply untwisted the strands, and so divided it into two ropes, each of the original length, but one-half the thickness. He would thus be able to tie the two together and make a rope nearly twice the original length, with which it is quite conceivable that he made good his escape from the dungeon.
50.—The Underground Maze
How did the jester find his way out of the maze in the dark? He had simply to grope his way to a wall and then keep on walking without once removing his left hand (or right hand) from the wall. Starting from A, the dotted line will make the route clear when he goes to the left. If the reader tries the route to the right in the same way he will be equally successful; in fact, the two routes unite and cover every part of the walls of the maze except those two detached parts on the left-hand side—one piece like a U, and the other like a distorted E. This rule will apply to the majority of mazes and puzzle gardens; but if the centre were enclosed by an isolated wall in the form of a split ring, the jester would simply have gone round and round this ring.
See the article, "Mazes, and How to Thread Them," in A. in M.
51.—The Secret Lock
This puzzle entailed the finding of an English word of three letters, each letter being found on a different dial. Now, there is no English word composed of consonants alone, and the only vowel appearing anywhere on the dials is Y. No English word begins with Y and has the two other letters consonants, and all the words of three letters ending in Y (with two consonants) either begin with an S or have H, L, or R as their second letter. But these four consonants do not appear. Therefore Y must occur in the middle, and the only word that I can find is "PYX," and there can be little doubt that this was the word. At any rate, it solves our puzzle.
52.—Crossing the Moat
No doubt some of my readers will smile at the statement that a man in a boat on smooth water can pull himself across with the tiller rope! But it is a fact. If the jester had fastened the end of his rope to the stern of the boat and then, while standing in the bows, had given a series of violent jerks, the boat would have been propelled forward. This has often been put to a practical test, and it is said that a speed of two or three miles an hour may be attained. See W. W. Rouse Ball's Mathematical Recreations.
53.—The Royal Gardens
This puzzle must have struck many readers as being absolutely impossible. The jester said: "I had, of a truth, entered every one of the sixteen gardens once, and never more than once." If we follow the route shown in the accompanying diagram, we find that there is no difficulty in once entering all the gardens but one before reaching the last garden containing the exit B. The difficulty is to get into the garden with a star, because if we leave the B garden we are compelled to enter it a second time before escaping, and no garden may be entered twice. The trick consists in the fact that you may enter that starred garden without necessarily leaving the other. If, when the jester got to the gateway where the dotted line makes a sharp bend, his intention had been to hide in the starred garden, but after he had put one foot through the doorway, upon the star, he discovered it was a false alarm and withdrew, he could truly say: "I entered the starred garden, because I put my foot and part of my body in it; and I did not enter the other garden twice, because, after once going in I never left it until I made my exit at B." This is the only answer possible, and it was doubtless that which the jester intended.
See "The Languishing Maiden," in A. in M.
54.—Bridging the Ditch
The solution to this puzzle is best explained by the illustration. If he had placed his eight planks, in the manner shown, across the angle of the ditch, he would have been able to cross without much trouble. The king's jester might thus have well overcome all his difficulties and got safely away, as he has told us that he succeeded in doing.
THE SQUIRE'S CHRISTMAS PUZZLE PARTY
HOW THE VARIOUS TRICKS WERE DONE
The record of one of Squire Davidge's annual "Puzzle Parties," made by the old gentleman's young lady relative, who had often spent a merry Christmas at Stoke Courcy Hall, does not contain the solutions of the mysteries. So I will give my own answers to the puzzles and try to make them as clear as possible to those who may be more or less novices in such matters.
55.—The Three Teacups
Miss Charity Lockyer clearly must have had a trick up her sleeve, and I think it highly probable that it was conceived on the following lines. She proposed that ten lumps of sugar should be placed in three teacups, so that there should be an odd number of lumps in every cup. The illustration perhaps shows Miss Charity's answer, and the figures on the cups indicate the number of lumps that have been separately placed in them. By placing the cup that holds one lump inside the one that holds two lumps, it can be correctly stated that every cup contains an odd number of lumps. One cup holds seven lumps, another holds one lump, while the third cup holds three lumps. It is evident that if a cup contains another cup it also contains the contents of that second cup.
There are in all fifteen different solutions to this puzzle. Here they are:—
The first two numbers in a triplet represent respectively the number of lumps to be placed in the inner and outer of the two cups that are placed one inside the other. It will be noted that the outer cup of the pair may itself be empty.
56.—The Eleven Pennies
It is rather evident that the trick in this puzzle was as follows:—From the eleven coins take five; then add four (to those already taken away) and you leave nine—in the second heap of those removed!
57.—The Christmas Geese
Farmer Rouse sent exactly 101 geese to market. Jabez first sold Mr. Jasper Tyler half of the flock and half a goose over (that is, 50-½ + ½, or 51 geese, leaving 50); he then sold Farmer Avent a third of what remained and a third of a goose over (that is, 16-2/3 + 1/3, or 17 geese, leaving 33); he then sold Widow Foster a quarter of what remained and three-quarters of a goose over (that is, 8-1/4 + 3/4 or 9 geese, leaving 24); he next sold Ned Collier a fifth of what he had left and gave him a fifth of a goose "for the missus" (that is, 4-4/5 + 1/5 or 5 geese, leaving 19). He then took these 19 back to his master.
58.—The Chalked Numbers
This little jest on the part of Major Trenchard is another trick puzzle, and the face of the roguish boy on the extreme right, with the figure 9 on his back, showed clearly that he was in the secret, whatever that secret might be. I have no doubt (bearing in mind the Major's hint as to the numbers being "properly regarded") that his answer was that depicted in the illustration, where boy No. 9 stands on his head and so converts his number into 6. This makes the total 36—an even number—and by making boys 3 and 4 change places with 7 and 8, we get 1278 and 5346, the figures of which, in each case, add up to 18. There are just three other ways in which the boys may be grouped: 1368—2457, 1467—2358, and 2367—1458.
59.—Tasting the Plum Puddings
The diagram will show how this puzzle is to be solved. It is the only way within the conditions laid down. Starting at the pudding with holly at the top left-hand corner, we strike out all the puddings in twenty-one straight strokes, taste the steaming hot pudding at the end of the tenth stroke, and end at the second sprig of holly.
Here we have an example of a chess rook's path that is not re-entrant, but between two squares that are at the greatest possible distance from one another. For if it were desired to move, under the condition of visiting every square once and once only, from one corner square to the other corner square on the same diagonal, the feat is impossible.
There are a good many different routes for passing from one sprig of holly to the other in the smallest possible number of moves—twenty-one—but I have not counted them. I have recorded fourteen of these, and possibly there are more. Any one of these would serve our purpose, except for the condition that the tenth stroke shall end at the steaming hot pudding. This was introduced to stop a plurality of solutions—called by the maker of chess problems "cooks." I am not aware of more than one solution to this puzzle; but as I may not have recorded all the tours, I cannot make a positive statement on the point at the time of writing.
60.—Under the Mistletoe Bough
Everybody was found to have kissed everybody else once under the mistletoe, with the following additions and exceptions: No male kissed a male; no man kissed a married woman except his own wife; all the bachelors and boys kissed all the maidens and girls twice; the widower did not kiss anybody, and the widows did not kiss each other. Every kiss was returned, and the double performance was to count as one kiss. In making a list of the company, we can leave out the widower altogether, because he took no part in the osculatory exercise.
Now, if every one of these 39 persons kissed everybody else once, the number of kisses would be 741; and if the 12 bachelors and boys each kissed the 10 maidens and girls once again, we must add 120, making a total of 861 kisses. But as no married man kissed a married woman other than his own wife, we must deduct 42 kisses; as no male kissed another male, we must deduct 171 kisses; and as no widow kissed another widow, we must deduct 3 kisses. We have, therefore, to deduct 42+171+3=216 kisses from the above total of 861, and the result, 645, represents exactly the number of kisses that were actually given under the mistletoe bough.
61.—The Silver Cubes
There is no limit to the number of different dimensions that will give two cubes whose sum shall be exactly seventeen cubic inches. Here is the answer in the smallest possible numbers. One of the silver cubes must measure 2-23278/40831 inches along each edge, and the other must measure 11663/40831 inch. If the reader likes to undertake the task of cubing each number (that is, multiply each number twice by itself), he will find that when added together the contents exactly equal seventeen cubic inches. See also No. 20, "The Puzzle of the Doctor of Physic."
THE ADVENTURES OF THE PUZZLE CLUB
62.—The Ambiguous Photograph
One by one the members of the Club succeeded in discovering the key to the mystery of the Ambiguous Photograph, except Churton, who was at length persuaded to "give it up." Herbert Baynes then pointed out to him that the coat that Lord Marksford was carrying over his arm was a lady's coat, because the buttons are on the left side, whereas a man's coat always has the buttons on the right-hand side. Lord Marksford would not be likely to walk about the streets of Paris with a lady's coat over his arm unless he was accompanying the owner. He was therefore walking with the lady.
As they were talking a waiter brought a telegram to Baynes.
"Here you are," he said, after reading the message. "A wire from Dovey: 'Don't bother about photo. Find lady was the gentleman's sister, passing through Paris.' That settles it. You might notice that the lady was lightly clad, and therefore the coat might well be hers. But it is clear that the rain was only a sudden shower, and no doubt they were close to their destination, and she did not think it worth while to put the coat on."
63.—The Cornish Cliff Mystery
Melville's explanation of the Cornish Cliff Mystery was very simple when he gave it. Yet it was an ingenious trick that the two criminals adopted, and it would have completely succeeded had not our friends from the Puzzle Club accidentally appeared on the scene. This is what happened: When Lamson and Marsh reached the stile, Marsh alone walked to the top of the cliff, with Lamson's larger boots in his hands. Arrived at the edge of the cliff, he changed the boots and walked backwards to the stile, carrying his own boots.
This little manœuvre accounts for the smaller footprints showing a deeper impression at the heel, and the larger prints a deeper impression at the toe; for a man will walk more heavily on his heels when going forward, but will make a deeper impression with the toes in walking backwards. It will also account for the fact that the large footprints were sometimes impressed over the smaller ones, but never the reverse; also for the circumstance that the larger footprints showed a shorter stride, for a man will necessarily take a smaller stride when walking backwards. The pocket-book was intentionally dropped, to lead the police to discover the footprints, and so be put on the wrong scent.
64.—The Runaway Motor-Car
Russell found that there are just twelve five-figure numbers that have the peculiarity that the first two figures multiplied by the last three—all the figures being different, and there being no 0—will produce a number with exactly the same five figures, in a different order. But only one of these twelve begins with a 1—namely, 14926. Now, if we multiply 14 by 926, the result is 12964, which contains the same five figures. The number of the motor-car was therefore 14926.
Here are the other eleven numbers:—24651, 42678, 51246, 57834, 75231, 78624, 87435, 72936, 65281, 65983, and 86251.
Compare with the problems in "Digital Puzzles," section of A. in M., and with Nos. 93 and 101 in these pages.
65.—The Mystery of Ravensdene Park
The diagrams show that there are two different ways in which the routes of the various persons involved in the Ravensdene Mystery may be traced, without any path ever crossing another. It depends whether the butler, E, went to the north or the south of the gamekeeper's cottage, and the gamekeeper, A, went to the south or the north of the hall. But it will be found that the only persons who could have approached Mr. Cyril Hastings without crossing a path were the butler, E, and the man, C. It was, however, a fact that the butler retired to bed five minutes before midnight, whereas Mr. Hastings did not leave his friend's house until midnight. Therefore the criminal must have been the man who entered the park at C.
66.—The Buried Treasure
The field must have contained between 179 and 180 acres—to be more exact, 179.37254 acres. Had the measurements been 3, 2, and 4 furlongs respectively from successive corners, then the field would have been 209.70537 acres in area.
One method of solving this problem is as follows. Find the area of triangle APB in terms of x, the side of the square. Double the result=xy. Divide by x and then square, and we have the value of y2 in terms of x. Similarly find value of z2 in terms of x; then solve the equation y2+z2=32, which will come out in the form x4-20x2=-37. Therefore x2=10+(sqrt{63})=17.937254 square furlongs, very nearly, and as there are ten acres in one square furlong, this equals 179.37254 acres. If we take the negative root of the equation, we get the area of the field as 20.62746 acres, in which case the treasure would have been buried outside the field, as in Diagram 2. But this solution is excluded by the condition that the treasure was buried in the field. The words were, "The document … states clearly that the field is square, and that the treasure is buried in it."
